Loading
Notes
Study Reminders
Support
Text Version

Stress, Strain and Strain Rate and Shear Plane Angle

Set your study reminders

We will email you at these times to remind you to study.
  • Monday

    -

    7am

    +

    Tuesday

    -

    7am

    +

    Wednesday

    -

    7am

    +

    Thursday

    -

    7am

    +

    Friday

    -

    7am

    +

    Saturday

    -

    7am

    +

    Sunday

    -

    7am

    +

Manufacturing Processes - Casting and Joining
Prof. Sounak Kumar Choudhury
Department of Mechanical Engineering
Indian Institute of Technology Kanpur
Lecture - 08
Stress, Strain and Strain Rate and Shear Plane Angle
Hello and welcome back to the course on Manufacturing Processes - Casting and
Joining. Let me remind you that in the casting, we have already discussed the total
solidification time - how to find it out, we have discussed how the solidification goes in a
pure metal and in a in alloys.
So, those things should be now clear to you. Then, we started discussing about the
application of the theory part whatever we have learnt including the Chvorinov’s rule
which says that the TST, total solidification time is,
2
m
V TST C
A
  =     .
(Refer Slide Time: 01:10)
This is a direct application of the Chvorinov’s rule in practice through the numerical
example. It says that a riser had to be designed in the shape of a sphere, that is for the
sand mold casting and the casting has to be a rectangular plate with 200 millimeter
length, 100 millimeter width and the 18 millimeter thickness.The total solidification time is given which is 3.5 minute, we have to determine the
diameter of the riser so that it will take 25 percent longer for the riser to solidify. Before
we go into details of this numerical example, I would like to clarify this Chvorinov’s rule
- what it actually identifies is that the riser design we can make according to Chvorinov’s
rule.
It is already said that it depends on the V
A
     ratio and we can design in such a way that
the the liquefied material in the molten metal inside the riser solidifies later than the
molten metal inside the mold cavity. Then only we can ensure the feeding of the molten
metal to the shrinkages that happens during the solidification.
Now, coming back to the numerical example, casting rectangular plate sizes are given;
length, width and thickness, we can multiply them and find out the casting volume. So,
the casting volume can be found out which is equal to 360,000 3 mm . Similarly, the area
can be also found out because it is a rectangular plate with a length, width and thickness
given. this is becoming 50800 2
mm . So, the volume by area ratio of the rectangular
plate is 7.0866.
Now, according to the Chvorinov’s rule, the total solidification time, therefore, since we
know the V
A
     , this will be equal to the Cm which is a mold constant into the
2
V
A
      and
this is given which is 3.5 minute. Let us say it is the total solidification time of such kind
of a rectangular plate already known in practice and it is given as 3.5 minute.
From this equation, we can find out the value of the Cm which is 3.5 minute divided by
(7.0866)
2
. This is in 2 min/ mm . This is 7.0866, it is in 2
mm . Now, the riser volume;
riser is given as the sphere.
So, the riser volume is
3
6
π D . This is 0.5236 3 D , π is constant. The D, we can actually
put it in here 6
π is 0.5236. So, 6.5236D3 is actually the riser volume. The riser area is
2 π D and this is a sphere. So, it is 3.1416 which is 2 π D . The volume by area of the riser will be 0.5236 D3
divided
by 3.1416 2 D , which gives you 0.1167 D . Now, the total solidification time in the riser ,
this will be 1.25 times more because it is 25 percent longer; longer than the casting TST
which is 3.5 minute.
So, we are finding out the total solidification time of the riser and here it is given that we
have to design the riser in such a way that it takes 25 percent longer time for the riser to
solidify. Therefore, the total solidification time of the riser will be 1.25 into 3.5 we have
found out for the casting total solidification time, which gives you 4.375 minute,
Because we found out Cm is equal to 0.0697. This is the Cm and this is the V
A
      of the
riser because we are finding out the total solidification time of the riser. Therefore, we
will take this value V
A
      of the riser square and Cm we already know; this is 0.0697.
Why we are taking this Cm ? You understand this, I already said that Cm of the riser and
the casting will be the same.
We found it out for the casting. So, this can be taken as 0.0697. This will give you
0.001936 2 D and from here D is equal to we can find out root over 2259.7. This will be
47.5 mm . This as you can see is quite simple; based on the basic principle that the total
solidification time is equal to mold constant into
2
V
A
      .
This is the Chvorinov’s rule and accordingly, we can find out the V
A
      of the casting
V
A
      of the riser and then, we can solve this because in one initially we can find out the
Cm and that Cm will be the same for the riser. So, put that value and find out what is the
total solidification time. From there, you can find out the dimension. This is how it is
solved. (Refer Slide Time: 08:07)
Now, the second example based on the same principle, that is the Chvorinov’s rule. It
says, a cylindrical riser is to be used for a sand-casting mold. For a given cylinder
volume, determine the diameter to length ratio that will maximize the time to solidify.
This question is little tricky, little different than the earlier one.
Because here, it is only the conditions are given; that you have to design a cylindrical
riser for the sand casting and for that, you have to determine the diameter to length ratio
that will maximize the time to solidify. Let us see how to solve this.
Now to maximize total solidification time, the V
A
     ratio must be maximum because the
total solidification time is directly proportional to the V
A
     ;
2
V
A
      . Now, we have a
cylinder. The cylinder volume is
2
4
D L   π
   
. From here, we can find out L as 2
4V
π D .
Similarly, cylinder area is
2
2.
4
D DL π
+π . Substitute the expression for L that we found
out from cylinder volume, which is 2
4V
π D . Put this in here, from the volume equation in
the area equation; this is the area equation. A is equal to
2
2.
4
D DL π
+π .This gives you that A is equal to
2
2
D DL π
+π and instead of L, we are putting the value
of the 2
4V
π D . Ultimately, it will give you 4V
D . So, all together it will be
2 4
2
D V
D
π
+ which is equal to the area.
Now differentiate the area equation with respect to D. Now, this will be dA first
derivative of the A which is this with respect to capital D; this will be, 2
dA V4 D
dD D = − π if
you take the first derivative of this equation, and that will be equal to 0. That is, the first
derivative is equal to 0.
If we rearrange that, it will be π D is equal to we will take it to the right side, 2
4V
D . From
here, we can find out the D value. D will be
0.333 4V
π
      ; Now, from the previous
expression for L, substitute in the equation for D that we have right now developed. L is
equal to 2
4V
π D . Initially, from the cylinder volume, we found out that.
In here, we can put the value of the D, that D we just now found out from the first
derivative of the area with respect to the capital D ; putting the value of D and after
rearranging as shown in the slide, we can find out the value of L as
0.333 4V
π
      , which is
the same as the value of L.
Thus, the optimum values are D is equal to L, you can see that the D is equal to
0.333 4V
π
      and we got the value of the L the same. So, the D and the L are the same which
is
0.333 4V
π
      . Therefore, this is the optimum D
L
     ratio because we have taken the first
derivative is equal to 0 and this will be equal to 1 because D is equal to L. That is how we can find out the optimum design of the D
L
     . As you can see that this
also looks like a tricky example; but as it is, not so and it is simple, you are using again,
the Chvorinov’s rule and then, you are finding out the volume and the area of the casting
and the riser subsequently.
(Refer Slide Time: 13:35)
Probably to clarify the entire theory little more, we will discuss some more examples
and the third example says that a cylindrical riser, again the riser is cylindrical of 6
centimeter diameter and 6 centimeter height has to be designed. Now, let me tell you in
between that why we are concentrating more on the riser design. I will once again repeat
that this is the most important factor that we have to get the casting with proper
dimensions, with proper accuracy as far as possible.
Now, you cannot avoid the shrinkage process because this is the metallurgical property
of the material; Molten metal solidifies in three different stages, it will have the
contraction and therefore, it will have the shrinkage. So, it is very important for us that
during the production of the casting, during the fabrication of the casting, we have to
compensate for the shrinkages.
And only way to compensate for that is to feed extra molten material to the shrinkages,
to the solidified metal inside the mold cavity so that those shrinkages can be to some
extent, to the larger extent compensated for. That is why the design of the riser is very important that the total solidification time of the riser should be more than the total
solidification time of the casting.
Because casting should solidify fast, faster than the riser and therefore, the total
solidification time of the riser should be moreso that it solidifies at a slower rate than the
casting. That is why in all examples we are giving importance to the riser design; in
practice also we will see the riser design will be one of the most important factors,
depending on the kind of casting that you are making.
Because the riser design will be very different if the design of the casting is different.
All the time, we have to consider the V
A
     ratio of the casting and depending on that ratio
of the casting, we have to design the volume and area ratio of the riser .
A cylindrical riser of 6 centimeter diameter and 6 centimeter height has to be designed
for a sand casting mould. So, producing a steel rectangular plate, again the plate is
rectangular casting. This is of 7 centimeter into 10 centimeter into 2 centimeter height.
This is the length, this is the width, this is the height; having the total solidification time
of 1.36 minute.
Now, here also you can see the total solidification time is given like in one of the earlier
examples because this is taken from the earlier practice. Like if we have made a similar
casting somewhere else, so you know what is the solidification time of that if the
material is the same and if the dimensions are the same.
This is known, let us say, therefore, determine the total solidification time for the riser.
Now, here the plate is given. This is the rectangular plate. 7 x 10 x 2 will be the volume
of the plate in 3 cm . Volume of the plate is (7 x 10 x 2) which is 140 3 cm . Now, the area
of the plate will be, we have done it earlier, 2(7x10 + 7x2 + 10x2). This will be 208
2
cm .
Now, the V
A
      of the casting, this will be 140 divided by this 208. this is coming out to
be 0.673, unit less , it is a ratio. Now, the mould constant. By the way the mould constant always comes out as minute by centimeter square. This is the mould constant which is
total solidification time divided by
2
V
A
      .
Total solidification time is known which is 1.36 minute divided by V
A
     ,already we
found out, which is (0.673)
2
. This is coming out to be 0.453. 1.36
0.453
     ,which is 3
2 min/ cm . This is the mould constant. Now, for the cylindrical riser , the volume of the
riser is
2
4
D h π ,h is the height. Area of the riser, A is
2
2.
4
D Dh π
π
 
+    
, this also we
have done it earlier that how to find out the volume and the area.
So, the D
h
     is equal to 1 . Therefore, the V is equal to
3
4
π D . Now, the area of the riser
we have found out to be
2
2 2.
4
D D π
π
 
+    
. From here, you put these values which will be
2 1.5π D , if we simplify this. Therefore, the V
A
      of the riser because we have the V riser
and we have the A riser, we take the ratio, it will be
6
  D
    and
6
  D
   is 1 cm , since D = 6
cm given. .
Now, the total solidification time is
2
3.
6
  D
    because Cm is equal to 3. We took the Cm
here and
6
  D
   is the V
A
      of riser square. This is equal to 1 and therefore, it will be 3
minutes. The total solidification time, that is what we had to determine, what is the total
solidification time for the riser, this has come out to be 3 minutes.(Refer Slide Time: 20:45)
The next numerical example that we will discuss is the following. The two castings of
the same metal have the same surface area; one casting is in the form of a sphere and the
other is a cube. What is the ratio of the solidification time for the sphere to that of the
cube? See if we look at this numerical example, no data is given except that it is sphere
and cube. Knowing very well that we can find out the volume of the sphere, volume of
the cube, area of the cube, area of the square and so on. Let us see. It is an interesting
question, interesting example.
Now, the solution is the total solidification time ts, this is the mould constant multiplied
by
2
V
A
      which is Chvorinov’s rule. B is the mould constant. Well, earlier if we
designated that as the Cm ; does not matter, here we are saying suppose it is B, this is the
mould constant again equivalent to that Cm . Now, for sphere, one casting is in the form
of a sphere.
So, for the sphere of radius r,
2
V
A
      ; V sphere is 4 3
3
π r is the formula for the volume of a
sphere; when the r is the radius of the sphere. Now, that is divided by the A, area of the
sphere and area of the sphere is 2 4π r ; r is again the radius of the sphere. So, V
A
     ,that is volume by area ratio will be
2
9
r . Now, for cube, we have in the form of
a sphere one casting and another is in the cube; but both of them have the same surface
area. You mind it, this is the condition for this example. Now, let us find out for the
cube. Let us say the cube side is l.
Volume of the cube V is 3
l and the area is the 2 6l . the ratio is
3
2 6
l
l
which is. All
together,
2
V
A
      will give you l square divided by 36 because it will be l by 6 inside and
square of that will be
2
36
l .
Once again, we have found out for the sphere what is the V
A
     ratio, this is clear.
Because the volume of a sphere is 4 3
3
π r and the area of the sphere is 2 4π r . So, the ratio
will be
2
9
r . Now, for cube, the volume of the cube is l
3. Let us say side is l; small l and
the area of cube is 6 l
2 .
So, their ratio will be
3
2 6
l
l
and the
2
V
A
      as it is used in the Chvorinov’s rule will be
2
36
l ,
this is what has been said. Now, from Chvorinov’s rule, the ratio of tsphere and tcube that is
the total solidification time of the sphere and the total solidification time of the cube will
be equal to these two ratios because the constant is getting cancelled, it is the same.
So, it will be
2
9
r which is for the sphere volume by area ratio square divided by the
volume by area ratio square of the cube. This is
2
36
l and this will give you the
2
2
4r
l
this
you are getting from the Chvorinov’s rule, that what is the ratio of the total solidification
time of the sphere and the total solidification time of the cube.
Because we found out separately; we can find out the total solidification time here, as
you can see the B constant, that is the mould constant is not coming into picture because it is getting cancelled. Therefore, this value is not given and this value is not required
actually because this is getting cancelled and we already said that the mould constant for
the casting and mould constant of the riser will be the same.
Why it is same? Because the same material is going into the riser, same molten metal
which is fed to the cavity mould cavity, it goes into the riser. Therefore, the Cm is same
for the casting and for the riser. In this example very interesting is that mould constant is
not actually taken into consideration at all and while taking the ratio of the total
solidification time of the sphere and the cube, we are directly taking the
2
V
A
      of both of
them that is the sphere and the cube separately taken.
See how interesting it is and you can find out the surface area of the same. Therefore,
2 4π r is equal to 2 6l , from here it will be coming. Now, because the surface area we
know; surface area of this 2 4π r and here for the cube it is 2 6l . They are the same;
2 4π r , area of the sphere equal to area of the cube.
So, from here, you can simplify this and you can find out the ratio of sphere
cube
t
t , this will be
6
π
which is 1.91 because that
2
2
4r
l , this will be 6
π
;
6
π
is equal to 1.91.
The ratio of the total solidification time of the sphere and the cube will be equal to 1.91.
As you can see that the examples are simple, this is how you can understand the exact
application of the theory that we have gone through, particularly, as far as the riser
design is concerned and riser design is very important to get the proper kind of a casting.
So that the casting could be defect free and by defect, we mean to say not all the other
defects which will be discussing at a later stage; but particularly, in the dimensions, in
the form of the dimensions, in the form of the shrinkages. To compensate for the
shrinkage, the design for the riser is extremely important and this is how we can take
care of that.(Refer Slide Time: 28:45)
I hope this is understood. Now, we will discuss the shrinkages. What kind of shrinkages
we have and why these shrinkages happen and how those shrinkages can be taken care
of. Let us see this. Shrinkage occurs in three stages. I will remind you that the first stage
will be the liquid contraction during the cooling period to the solidification.
This is the cooling period, when it is starting or it started cooling. Contraction during the
phase change from liquid to solid is called the solidification shrinkage and the thermal
contraction of the solidified casting happens during the cooling to the room temperature.
So, these are the three stages when the shrinkage happens and we have to take care of the
shrinkage through the design of the riser, by the appropriate design of the riser. Once
again, it will be liquid contraction during the cooling prior to the solidification. Just
solidification starting at that point prior to solidification during the cooling, there will be
a contraction.
During the phase change from liquid to solid, it is getting changed from liquid to solid,
there will be contraction, there will be shrinkage; and the thermal contraction of the
solidified casting happens during the cooling to the room temperature, that is the last
stage. If you remember in that curve, there the shrinkage again takes place.
So, in these three stages the total shrinkages will be taking place that have to be taken
care of. The rest of the material, we will discuss in our next session of discussion.Thank you.